Question: $\overline{AB} = \sqrt{109}$ $\overline{AC} = {?}$ $A$ $C$ $B$ $\sqrt{109}$ $?$ $ \sin( \angle BAC ) = \frac{10\sqrt{109} }{109}, \cos( \angle BAC ) = \frac{3\sqrt{109} }{109}, \tan( \angle BAC ) = \dfrac{10}{3}$
Answer: $\overline{AB}$ is the hypotenuse $\overline{AC}$ is adjacent to $\angle BAC$ SOH CAH TOA We know the hypotenuse and need to solve for the adjacent side so we can use the cos function (CAH) $ \cos( \angle BAC ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\overline{AC}}{\overline{AB}}= \frac{\overline{AC}}{\sqrt{109}} $ $ \overline{AC}=\sqrt{109} \cdot \cos( \angle BAC ) = \sqrt{109} \cdot \frac{3\sqrt{109} }{109} = 3$